∏i=1n Piaicos2⁡ai=2019 where Pi represents the ith prime number (i.e., for example P1=2,P2=3,P3=5…… such that Pn<2019 and ai∈R . If elements of set A be solution of the equation represented by orderd n-tuple a1,a2,a3,….,an , then

2019=3×6732a1cos2⁡a13a2cos2⁡a25a3cos2⁡a3…..=3×673⇒a1=a3…..=0,a2cos2⁡a2=1 and a122cos2⁡a122=1a122=673xcos2⁡x=1 has 5 solutions in (−∞,9], infinite solutions in (−∞,∞), no solution in (−∞,0]

∏i=1n Piaicos2⁡ai=2019 where Pi represents the ith prime number (i.e., for example P1=2,P2=3,P3=5…… such that Pn<2019 and ai∈R . If elements of set A be solution of the equation represented by orderd n-tuple a1,a2,a3,….,an , then

2019=3×6732a1cos2⁡a13a2cos2⁡a25a3cos2⁡a3…..=3×673⇒a1=a3…..=0,a2cos2⁡a2=1 and a122cos2⁡a122=1a122=673xcos2⁡x=1 has 5 solutions in (−∞,9], infinite solutions in (−∞,∞), no solution in (−∞,0]

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$\prod_{i = 1}^{n} P_{i}^{a_{i} \cos^{2} a_{i}} = 2019$ where P_i represents the i^th prime number
(i.e., for example $P_{1} = 2, P_{2} = 3, P_{3} = 5 \dots \dots)$
such that $P_{n} < 2019 and a_{i} \in R$ . If elements of set A be solution of the equation represented by orderd n-tuple $(a_{1}, a_{2}, a_{3}, \dots ., a_{n})$ , then

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Number of elements in set A is finite

If $a_{i} \leq 0 \forall i = 1, 2, \dots, n$ , then A is null set

If $a_{i} \leq 9 \forall i = 1, 2, \dots ., n$ , then A is singleton set

If $a_{i} \leq 9 \forall i = 1, 2, \dots ., n$ , then number of element in set A is 25

answer is B, C.

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Detailed Solution

$\begin{array}{l} 2019 = 3 \times 673 \\ 2^{a_{1} \cos^{2} a_{1}} 3^{a_{2} \cos^{2} a_{2}} 5^{a_{3} \cos^{2} a_{3}} \dots . . = 3 \times 673 \\ \Rightarrow a_{1} = a_{3} \dots . . = 0, a_{2} \cos^{2} a_{2} = 1 and a_{122} \cos^{2} a_{122} = 1 (a_{122} = 673) \end{array}$
$x \cos^{2} x = 1 has 5 solutions in (- \infty, 9], infinite solutions in (- \infty, \infty),$ no solution in $(- \infty, 0]$